Question

If a room temperature roast cools from 68°F to 25°F in 5 hours at freezer temperature of 20°F,

Answer 1

Answer:

9 hours

Step-by-step explanation:

According to Newton's laws of cooling

dT/dt = -k(T - A)

Let U = T - A

dU/dT = 1

dU = dT

dt/dt = -kU

dT = -kU(dt)

dU/U = -kdt

On integration

ln(U) = -kt + C

U = Ce^-kt

T - A = Ce^-kt

T(0) = 68

T(5) = 25

68 - 20 = Ce^-k(0)

C = 48

and

25 -20 = 48e^-k(5)

5 = 48e^-5k

e^-5k = 5/48

-5k = ln (5/48)

k = - ln(5/48) / 5

k = - 0.4524

T - 20 = 48e^-0.4524t

When T = 21

21 -20 = 48e^-0.4524t

1 = 48e^-0..4524t

e^-0.4524t = 1/48

-0.4524t = ln (1/48)

t = - ln(1/48) / 0.4524

t = 8.5570

t= 9 hours ( to the nearest hour)